Tully K. A Level Maths tutor, GCSE Maths tutor, Uni Admissions Test -...

Tully K.

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Engineering Science (Masters) - Oxford, Keble College University

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About me

Hi,

As an engineer, I think the power of maths is amazing, and I'd like to share some understanding of maths and science so that you can appreciate it and use it better. I like to break things down into understandable examples, sometimes what's written in the textbook just doesn't make sense!

At the end of the day an exam is graded on how many marks you can get, and a bit of exam technique can go a long way to ensuring you fulfill your potential. I have learned a lot about how to maximise exam marks and this will form an important part of my tutorials if it is something you wish to improve.

I look forward to meeting you and discussing what you want from our tutorials.

 

Tully

Hi,

As an engineer, I think the power of maths is amazing, and I'd like to share some understanding of maths and science so that you can appreciate it and use it better. I like to break things down into understandable examples, sometimes what's written in the textbook just doesn't make sense!

At the end of the day an exam is graded on how many marks you can get, and a bit of exam technique can go a long way to ensuring you fulfill your potential. I have learned a lot about how to maximise exam marks and this will form an important part of my tutorials if it is something you wish to improve.

I look forward to meeting you and discussing what you want from our tutorials.

 

Tully

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
ChemistryA-level (A2)A*
PhysicsA-level (A2)A*

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr
.PAT.Uni Admissions Test£25 /hr

Questions Tully has answered

What is 'Chain Rule' and why is it useful?

The chain rule is most commonly seen in Leibniz's notation, which is as follows:

 

dz/dx = dz/dy * dy/dx

 

You can remember it intuitively by thinking of the two 'dy' terms cancelling to leave dz/dx.

 

So why use the chain rule?

You are used to differentiating equations in the form y = f(x), but say both sides of the equation where functions eg g(y) = f(x) and you had to differentiate the equation with respect to x. 

g is a function of y, not x, so you can't simply calculate dg(y)/dx like you can df(x)/dx. Using the chain rule we can express dg(y)/dx as dg(y)/dy * dy/dx. These two terms can be calculated (assuming y is a function of x). This is really what the chain rule is saying: that the derivative of a function composition can be expressed as a product of the respective derivatives.

 

Another example of when the chain rule might come in useful is in mechanics: Acceleration is defined as the derivative of velocity: dv/dt. Sometimes though it might be useful to integrate acceleration of a distance, x, rather than over time. To eliminate time from this expression we can use the chain rule by saying dv/dt = dv/dx * dx/dt. Then noting that dx/dt is in fact velocity (v = dx/dt) we can write that dv/dt = v * dv/dx thus making acceleration a function only of velocity and position.

The chain rule is most commonly seen in Leibniz's notation, which is as follows:

 

dz/dx = dz/dy * dy/dx

 

You can remember it intuitively by thinking of the two 'dy' terms cancelling to leave dz/dx.

 

So why use the chain rule?

You are used to differentiating equations in the form y = f(x), but say both sides of the equation where functions eg g(y) = f(x) and you had to differentiate the equation with respect to x. 

g is a function of y, not x, so you can't simply calculate dg(y)/dx like you can df(x)/dx. Using the chain rule we can express dg(y)/dx as dg(y)/dy * dy/dx. These two terms can be calculated (assuming y is a function of x). This is really what the chain rule is saying: that the derivative of a function composition can be expressed as a product of the respective derivatives.

 

Another example of when the chain rule might come in useful is in mechanics: Acceleration is defined as the derivative of velocity: dv/dt. Sometimes though it might be useful to integrate acceleration of a distance, x, rather than over time. To eliminate time from this expression we can use the chain rule by saying dv/dt = dv/dx * dx/dt. Then noting that dx/dt is in fact velocity (v = dx/dt) we can write that dv/dt = v * dv/dx thus making acceleration a function only of velocity and position.

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3 years ago

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